Question

Differentiate

${\tan }^{-1}\left(\frac{4x}{1-4x^2}\right),-\frac{1}{2}<x<\frac{1}{2}$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left\{\frac{4x}{1-4x^2}\right\} \\ \mathrm{put}2x=\tan \theta \\ \Rightarrow y={\tan }^{-1}\left\{\frac{2\tan \theta }{1-{\tan }^2\theta }\right\} \\ \Rightarrow y={\tan }^{-1}\left(\tan 2\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Here},-\frac{1}{2}<x<\frac{1}{2} \\ \Rightarrow -1<2x<1 \\ \Rightarrow -1<\tan \theta <1 \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\theta <\frac{\mathrm{\pi }}{4} \\ \Rightarrow -\frac{\mathrm{\pi }}{2}<2\theta <\frac{\mathrm{\pi }}{2} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=2\theta \left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow y=2{\tan }^{-1}\left(2x\right)\left[\mathrm{Since},2x=\tan \theta \right] \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=2\left(\frac{1}{1+{\left(2x\right)}^2}\right)\frac{d}{dx}\left(2x\right) \\ \Rightarrow \frac{dy}{dx}=2\left(\frac{1}{1+4x^2}\right)\times 2 \\ \therefore \frac{dy}{dx}=\frac{4}{1+4x^2} \end{gathered}$$